7
$\begingroup$

Let $A$ and $B$ be two Hermitian complex matrices. (a) Prove that $\operatorname{tr}(AB)$ is real. (b) Prove that if $A, B$ are positive, then $\operatorname{tr}(AB)>0$.

(a) The trace of Hermitian matrix is a real number, since $a_{ii} = \bar{a}_{ii}$, that also means that all eigenvalues are real. I can't proceed to conclusion that $\operatorname{tr}(AB)$ is real, since $\operatorname{tr}(AB) \neq \operatorname{tr}A\cdot\operatorname{tr}B$ and product of two Hermitian matrices is also Hermitian only if these matrices commute, which is not the case for arbitrary Hermitian matrices.

(b) Am I missing something or the question is indeed so easy? If all entries $a_{ij}>0$ and $b_{ij}>0$, then all $c_{ij}=\sum_{m}a_{im}\cdot b_{mj}>0$ too, finally $\operatorname{tr}(AB)=\sum_{i}c_{ii}>0$.

$\endgroup$
5
  • 1
    $\begingroup$ Don't they mean positive definite when they just say positive? $\endgroup$
    – Nik Pronko
    Aug 15, 2018 at 20:57
  • $\begingroup$ @NikPronko, maybe there is a typo and there should be positive definite, but the task states just positive. $\endgroup$
    – Hasek
    Aug 15, 2018 at 20:59
  • 1
    $\begingroup$ in context of complex matrices, positive usually means positive definite... $\endgroup$
    – user251257
    Aug 15, 2018 at 21:00
  • $\begingroup$ @user251257, it makes sense since we can't compare complex numbers with zero. I didn't pay attention to this, thanks! $\endgroup$
    – Hasek
    Aug 15, 2018 at 21:02
  • 1
    $\begingroup$ As part (b) isn't actually addressed to, a hint: As $A$ is positive definite, it has a positive definite square root, say $R$. That is $A = R^2$. Now, $AB=R(RB)RR^{-1}$ is similar to $RBR$, which is positive definite. $\endgroup$
    – user251257
    Aug 15, 2018 at 21:13

2 Answers 2

Reset to default
13
$\begingroup$

Since $A^T = \overline{A}$ and using basic properties of trace :

$$ tr(AB) = tr((AB)^T) = tr(B^T A^T) = tr(\overline{B} \ \overline{A})= tr(\overline{A} \ \overline{B}) $$

Therefore :

$$ tr(AB) = tr(\overline{A} \ \overline{B}) = tr(\overline{AB}) = \overline{tr(AB)} $$

Finally $tr(AB) \in \mathbb R$.

$\endgroup$
0
3
$\begingroup$

You can't just say that because the trace is over the diagonal elements of $(AB)_{ii}$. Instead $\lambda = tr(AB)=\sum A_{ij}B_{ji}$ then $\lambda^* = tr(A^*B^*)=tr(A^TB^T)=\sum A^T_{ij}B^T_{ji} = \sum A_{ji}B_{ji}$ which is clearly $tr(AB)$ under renaming dummy indicies, so $\lambda=\lambda^*$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .